Boundary Value Problems in Electrostatics II

Friedrich Wilhelm Bessel

(1784 - 1846)

December 23, 2000

Contents

1 Laplace Equation in Spherical Coordinates 2

1.1 Legendre Equation and Polynomials . . . . . . . . . . . . . . . . . . . 4

1.2 Solution of Boundary Value Problems with Azimuthal Symmetry . . 12

1.2.1 Example: A Sphere With a Specified Potential . . . . . . . . . 13

1.2.2 Example: Hemispheres of Opposite Potential . . . . . . . . . . 14

1.2.3 Example: Potential of an Isolated Charge . . . . . . . . . . . . 17

1.3 Behavior of Fields in Conical Holes and Near Sharp Points . . . . . . 19

1.4 Associated Legendre Polynomials; Spherical Harmonics . . . . . . . . 23

1.5 The Addition Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 26

1.6 Expansion of the Green’s Function in Spherical Harmonics . . . . . . 30

2 Laplace Equation in Cylindrical Coordinates; Bessel Functions 33

2.1 Example I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

2.2 Example II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

2.3 B.V.P. on Large Cylinders . . . . . . . . . . . . . . . . . . . . . . . . 47

2.4 Green’s Function Expansion in Cylindrical Coordinates . . . . . . . . 48

1

Although this is a new chapter, we continue to do things begun in

the previous chapter. In particular, the first topic is the separation of

variable method in spherical polar coordinates.

1 Laplace Equation in Spherical Coordinates

The Laplacian operator in spherical coordinates is

∇2 =1

r

∂2

∂r2r +

1

r2 sin θ

∂

∂θsin θ

∂

∂θ+

1

r2 sin2 θ

∂2

∂φ2. (1)

This is also a coordinate system in which it is possible to find a solution

in the form of a product of three functions of a single variable each:

Φ(r, θ, φ) = R(r)P (θ)Q(φ) = U(r)P (θ)Q(φ)/r. (2)

Operate on Φ with ∇2, and set the result equal to zero to find

PQ

r

d2U

dr2+

UQ

r2 sin θ

d

dθ

(sin θ

dP

dθ

)+

UP

r3 sin2 θ

d2Q

dφ2= 0 (3)

Multiply by r3 sin2 θ/UPQ to find

r2 sin2 θ

U

d2U

dr2+

sin θ

P

d

dθ

(sin θ

dP

dθ

)+

1

Q

d2Q

dφ2= 0. (4)

The first two terms are independent of φ while the third depends only

on this variable. Thus the third must be a constant as must the sum

of the first two; the first of these conditions is

1

Q

d2Q

dφ2= C or

d2Q

dφ2= CQ, (5)

2

from which it follows that Q ∼ e√Cφ.

Now, a change in φ by 2π corresponds to no change whatsoever in

spatial position; therefore, we must have Q(φ + 2π) = Q(φ) because

a function describing a measurable quantity must be a single-valued

function of position. Hence we can conclude that√C = im where m

is an integer so that eim2π = 1. Thus C = −m2, and Q(φ)→ Qm(φ) =

eimφ, with m = 0,±1,±2,... . We recognize that the functions Qm can

be used to construct a Fourier series and are a complete orthogonal set

on the interval φ0 ≤ φ ≤ φ+ 2π.

Returning now to Laplace’s equation, Eq. (4), and using −m2 for

1Qd2Qdφ2 , we find

r2 sin2 θ

U

d2U

dr2+

sin θ

P

d

dθ

(sin θ

dP

dθ

)−m2 = 0 (6)

orr2

U

d2U

dr2+

1

P sin θ

d

dθ

(sin θ

dP

dθ

)− 1

sin2 θm2 = 0. (7)

In this expression we recognize that the first term depends only on r

and next two, only on θ, so we as usual conclude that each term must

be separately a constant and that the constants must add to zero. The

first equation extracted by this device is

d2U

dr2=A

r2U (8)

where A is the constant. It is a standard convention to write A as

l(l + 1) which is still quite general if l is allowed to be complex. Thus

3

the preceding equation becomes

d2U

dr2=l(l + 1)

r2U. (9)

The solutions of this ordinary, second-order, linear, differential equation

are two in number and are U ∼ rl+1 and U ∼ 1/rl. Before commenting

further on that, let us go on to the equation for P (θ).

1.1 Legendre Equation and Polynomials

Substitution of l(l + 1) for the first term in Eq. (7) produces

1

sin θ

d

dθ

(sin θ

dP

dθ

)+

l(l + 1)− m2

sin2 θ

P = 0. (10)

This is the generalized Legendre Equation; it is commonly written in

terms of a different variable, namely u ≡ cos θ. Then one has

d

dθ

(sin θ

dP

dθ

)=du

dθ

d

du

(√1− u2

du

dθ

dP

du

)= −√

1− u2d

du

(−(1− u2)

dP

du

)

(11)

and

l(l + 1)− m2

sin2 θ= l(l + 1)− m2

1− u2; (12)

hence,d

du

((1− u2)

dP

du

)+

l(l + 1)− m2

1− u2

P = 0 (13)

is the form of the generalized Legendre equation using u as the variable.

The interval of interest to us is 0 ≤ θ ≤ π which is −1 ≤ u ≤ 1.

We shall discuss first the special case of m = 0 which corresponds

to Q(φ) = 1, or a system for which Φ(x) is independent of φ; we shall

4

call such a potential azimuthally invariant; there are many interesting

systems which are more or less of this type. The equation for P is

d

du

((1− u2)

dP

du

)+ l(l + 1)P = 0, (14)

which is called the Legendre equation.

A standard procedure for solving this equation (and other similar

second-order differential equations) is to assume that the solution can

be written as a power series. Then there must be a smallest power α

in the series, so we can write

P (u) = uα∞∑

j=0

ajuj =

∞∑

j=0

ajuj+α (15)

from which we may evaluate the derivatives as

dP

du=∞∑

j=0

(j + α)ajuj+α−1, (16)

d2P

du2=∞∑

j=0

(j + α)(j + α− 1)ajuj+α−2, (17)

and

− d

duu2dP

du= −

∞∑

j=0

(j + α)(j + α + 1)ajuj+α. (18)

Substitution into the Legendre equation gives

∞∑

j=0

[(α + j)(α + j − 1)uα+j−2aj − (α + j)(α + j + 1)uα+jaj + l(l + 1)uα+jaj

]= 0,

(19)

or, if we shift the zero of j in each term so as to isolate individual

powers of u,

α(α− 1)uα−2a0 + α(α + 1)a1uα−1 +

5

∞∑

j=0

[(α + j + 2)(α + j + 1)aj+2 + (l(l + 1)− (α + j)(α + j + 1)) aj]uα+j = 0.(20)

The only way that this power series can vanish for all u on the

interval is to have the coefficient of each power of u vanish separately.

Thus I will list the coefficients:

j Coefficient of uα+j

−2 a0α(α− 1)

−1 a1α(α + 1)

j ≥ 0 [(α + j + 2)(α + j + 1)aj+2 + (l(l + 1)− (α + j)(α + j + 1)) aj]

The coefficient of the leading (smallest) power j = −2 is zero if α = 0

or 1; a0 = 0 is not an option because by definition the first term in the

expansion has a nonvanishing coefficient. Thus we find at this juncture

two possible allowed values of α.

α = 0, 1 (21)

In order that the coefficient of the next power j = −1 of u vanish we

must have either α = 0 or a1 = 0 (we can also have both); we can’t

have α = −1 because of our first condition. Finally, the condition that

the coefficient of uα+j vanish for j ≥ 0 is

aj+2 =(α + j)(α + j + 1)− l(l + 1)

(α + j + 1)(α + j + 2)aj. (22)

This is call the recurrence relation.

6

Consider this relation when j is very large, much larger than 1. In

this limit it simplifies to the statement that aj+2 = aj(1+O(1/j)) which

will produce a power series (for large powers) in the form of a sum of

terms proportional to u2j, all with the same coefficient. At u→ 1, this

sum will not converge, and so P is singular at u = 1. This is not an

allowed behavior for a solution to the Laplace equation, so we cannot

have such a function representing the potential. What must therefore

happen is that the series terminates which means that there must be

some j such that aj 6= 0 while aj+2 = 0. Examining Eq. (22), we see

that this j is such that

(α + j)(α + j + 1)− l(l + 1) = 0. (23)

This condition requires that α + j = l which is a condition on l; since

α is 0 or 1, and j is a non-negative integer, we see that l must be an

integer equal to or larger than α.

l ∈ Z l ≥ α (24)

Now, our recurrence relation gives us aj+2 from aj; hence, starting

from a0, we can get only the even coefficients aj, j even, and starting

from a1, we get the odd coefficients. Lets consider the odd series and

the even series separately. First consider the even series. Since at

termination of the series α + j = l, we see that l is even when α = 0

and l is odd when α = 1. Thus the even series terminates when

α = 0 and l even (25)

7

α = 1 and l odd

By similar arguments applied to the odd series, we can see that it

terminates when

α = 0 and l odd (26)

α = 1 and l even

Since l cannot be both odd and even, we can only have an even or

an odd series (they are actually equivalent). The other must be zero.

Since by convention, we choose a0 6= 0, it must be that the odd series

vanishes, and thus a1 = 0. Remembering that l is odd when α = 1 and

is even when α = 0, we see that the solutions are polynomials of degree

l. They are known as Legendre polynomials. It is easy to generate a few

of them, aside from normalization, starting from l = 0 and using the

recurrence relation. As for normalization, they are traditionally chosen

to be such that P (1) = 1. Let us add a subscript l to P to designate

the particular Legendre polynomial. The first few are

l Pl(u)

0 P0(u) = 1

1 P1(u) = u

2 P2(u) = 32u

2 − 12

3 P3(u) = 52u

3 − 32u

4 P4(u) = 358 u

4 − 154 u

2 + 38

8

There are other ways to generate the Legendre polynomials. For

example, one has Rodrigues’ formula which is

Pl(u) =1

2ll!

dl

dul(u2 − 1)l; (27)

it is easy to see that this generates a polynomial of degree l; one may

show that it is a solution to the Legendre equation by direct substitution

into that equation. Thus, it must be the Legendre polynomial (one

should also check normalization). If one expands the factor (u2 − 1)l

in Rodrigues’ formula using the binomial expansion and then takes the

derivatives, she/he will find that

Pl(u) =(−1)l

2l

l∑

n≥l/2

(−1)n(2n)!

n!(l − n)!(2n− l)!u2n−l (28)

Another way of generating Pl(u) is via the generating function

T (u, x) = (1− 2ux+ x2)−1/2 =∞∑

l=0

xlPl(u). (29)

If one takes l derivatives of this function with respect to x and then

sets x = 0, the result is l!Pl(u).

The Legendre functions have many properties that we will need to

make use of from time to time. For summaries of these, see e.g., ,

the section on Legendre functions in Abramowitz and Stegun starting

on p. 332 and also the section on orthogonal polynomials starting on

p. 771. Here we summarize some of the most significant properties.

First, orthogonality and normalization. Consider the integral

∫ 1

−1duPl(u)Pl′(u) =

1

2l+l′l!l′!

∫ 1

−1du

dl

dul(u2 − 1)l

dl′

dul′(u2 − 1)l

′. (30)

9

Suppose, without loss of generality, that l′ ≥ l and start by integrating

by parts,

∫ 1

−1duPl(u)Pl′(u) =

1

2l+l′l!l′!

dl

dul(u2 − 1)l

dl′−1

dul′−1(u2 − 1)l

′∣∣∣∣∣∣

1

−1

−∫ 1

−1du

dl+1

dul+1(u2 − 1)l

dl′−1

dul′−1(u2 − 1)l

′

.(31)

The first term in brackets vanishes because (u2 − 1) is zero at the end

points of the interval. Continuing from the right-hand side, integrate

in like fashion l′ − 1 more times. The result is

∫ 1

−1duPl(u)Pl′(u) = (−1)l

′ 1

2l+l′l!l′!

∫ 1

−1dudl+l

′(u2 − 1)l

dul+l′(u2 − 1)l

′. (32)

Now, dl+l′(u2− 1)l/dul+l

′= 0 if l′ > l, and so the integral is zero in this

case. If l′ = l, we have

∫ 1

−1duPl(u)Pl(u) = (−1)l

(2l)!

22l(l!)2

∫ 1

−1du(u2 − 1)l

= (−1)l(2l)!

22l(l!)22(−1)l

(2l)!!

(2l + 1)!!= 2

(2ll!)2(2l − 1)!!

22l(l!)2(2l + 1)!!=

2

2l + 1. (33)

Thus we have derived the relation

∫ 1

−1duPl(u)Pl′(u) =

2

2l + 1δll′ (34)

which expresses the orthogonality and normalization of the Legendre

polynomials.

Consider next recurrence relations. These provide, among other

things, a good way to generate values of Legendre polynomials on

10

computers. A number of recurrence relations can be derived using

Rodrigues’ formula and the Legendre equation. Consider, for example,

dPl+1

du=

1

2l+1(l + 1)!

dl+2

dul+2(u2 − 1)l+1

=l + 1

2l(l + 1)!

dl+1

dul+1

((u2 − 1)lu

)

=1

2ll!

dl

dul[(u2 − 1)l + 2lu2(u2 − 1)l−1

]

=1

2ll!

dl

dul[(u2 − 1)l + 2l(u2 − 1)l + 2l(u2 − 1)l−1

]

= (2l + 1)Pl(u) +dPl−1(u)

du, (35)

ordPl+1(u)

du− (2l + 1)Pl(u)− dPl−1(u)

du= 0. (36)

From this relation and the Legendre equation

d

du

((1− u2)

dPldu

)+ l(l + 1)Pl = 0 (37)

one may derive additional standard recurrence relations for the Legen-

dre polynomials. Several of these are

(l + 1)Pl+1 − u(2l + 1)Pl + lPl−1 = 0

(1− u2)dPldu

+ luPl − lPl−1 = 0

dPl+1

du− udPl

du− (l + 1)Pl = 0. (38)

These may be used to advantage in numerous applications such as doing

integrals of products of two Legendre polynomials and a power of u.

11

1.2 Solution of Boundary Value Problems with Azimuthal

Symmetry

Using what we have learned in the previous two sections, we are now

in a position to construct a general solution to the Laplace equation in

spherical coordinates under conditions of azimuthal invariance, that is,

when Φ(x) is independent of φ. The most general form that a solution

can have is

Φ(r, θ) =∞∑

l=0

(Alr

l +Blr−(l+1)

)Pl(cos θ). (39)

The Legendre polynomials form a complete set on the interval −1 ≤u ≤ 1 or 0 ≤ cos θ ≤ π. Thus any specified φ-independent potential

on a spherical surface can be expressed as a sum of Pl’s. If the volume

in which a solution is to be found includes the origin, then none of the

terms ∼ r−(l+1) can be included in the sum as they are singular at the

origin, and the potential will not be singular there. Similarly, if the

volume extends to r → ∞, then no terms ∼ rl are allowed. In the

former case, the conclusion is that Bl = 0 for all l, and in the latter

case, all Al = 0.

We consider now some examples.

12

1.2.1 Example: A Sphere With a Specified Potential

An isolated sphere of radius a is centered at the origin. By unspecified

means, the potential on its surface is maintained at

Φ(a, θ, φ) = Vo cos3(θ)

where θ is the polar angle. Find Φ(r, θ, φ) for all r > a.

This problem is azimuthally symmetric. Thus, in general,

Φ(r, θ) =∞∑

l=0

(Alr

l +Blr−(l+1)

)Pl(cos θ).

Since our volume contains all r > a, physics demands that Al = 0 for

all l. The constants Bl are then determined by matching the terms in

the series the boundary condition on the surface of the sphere. Recall

that

P0(x) 1

P1(x) x

P2(x) 12(3x2 − 1)

P3(x) 12(5x3 − 3x)

P4(x) 18(35x4 − 30x2 + 3)

P5(x) 18(63x5 − 70x3 + 15x)

So that,

Φ(a, θ, φ) = Vo cos3(θ) = Vo

(2

5P3(cos(θ)) +

3

5P1(cos(θ))

)

13

Thus

Φ(r, θ, φ) = Vo

2

5

(a

r

)4

P3(cos(θ)) +3

5

(a

r

)2

P1(cos(θ))

.

1.2.2 Example: Hemispheres of Opposite Potential

For the first, suppose that we need to solve the Laplace equation in-

side of a sphere of radius a given that on the surface, the potential is

specified as follows:

Φ(a, θ) =

V, 0 ≤ θ ≤ π/2

−V, π/2 ≤ θ ≤ π.(40)

VΦ =

-VΦ =

Then the expansion must take the form

Φ(r, θ) = V∞∑

l=0

Al

(r

a

)lPl(cos θ). (41)

Notice the introduction of the factor V on the right-hand side, along

with the use of the powers of a in the sum. These are included for

convenience. The scale of the potential and hence the size of a leading

term in the sum is set by V which also gives the correct dimensions

to the terms in the sum; it is thus natural to put this factor in each

14

term. The powers of a are included for the same reasons; r is of order

a and has the same dimensions so that leading coefficients Al are of

order unity and have dimension unity.

On the spherical surface, we have

Φ(a, θ) = V∞∑

l=0

AlPl(cos θ). (42)

In order to find a given coefficient An, we multiply this equation by

Pn(cos θ) and integrate over cos θ, recalling that d cos θ = − sin θdθ.

Making use of the orthogonality and normalization of the Legendre

polynomials, we find

∫ π0dθΦ(a, θ)Pn(cos θ) sin θ = V

∞∑

l=0

Al

(2

2l + 1

)δln = V

(2

2n+ 1

)An,

(43)

or

An =2n+ 1

2

[∫ π/20

dθPn(cos θ) sin θ −∫ ππ/2

dθPn(cos θ) sin θ

]

=2n+ 1

2

[∫ 1

0duPn(u)−

∫ 0

−1duPn(u)

](44)

Now use the inversion property of the Legendre polynomials, Pn(u) =

(−1)nPn(−u) to conclude that

An =2n+ 1

2

[∫ 1

0duPn(u)− (−1)n

∫ 1

0duPn(u)

]=

0 n even

(2n+ 1)∫ 10 duPn(u) n odd.

(45)

To complete the integral for the case of odd n we use a recurrence

15

relationdPn+1

du= (2n+ 1)Pn +

dPn−1

du(46)

so that:

An = (2n+1)∫ 1

0duPn(u) =

∫ 1

0du

[dPn+1

du− dPn−1

du

]= Pn−1(0)−Pn+1(0)

(47)

where we make use of the fact that Pn(1) = 1, independent of n. Fur-

ther, for even l,

Pl(0) = (−1)l/2(l − 1)!!

l!!(48)

where the “double factorial” sign means l!! = l(l − 2)(l − 4)...(2 or 1).

Hence

Pn−1(0)− Pn+1(0) = (−1)(n−1)/2

(n− 2)!!

(n− 1)!!+

n!!

(n+ 1)!!

= (−1)(n−1)/2 (n− 2)!!

(n+ 1)!!(n+ 1− n) = (−1)(n−1)/2 (n− 2)!!

(n+ 1)!!(2n+ 1). (49)

Now set n = 2m+ 1, m = 0,1,2,..., and have

Φ(r, θ) = V∞∑

m=0

Bm

(r

a

)2m+1

P2m+1(cos θ) (50)

where

Bm = (−1)m(2m− 1)!!

(2m+ 2)!!(4m+ 3). (51)

The first few terms in the expansion are

Φ(r, θ) =3

2V

r

aP1(cos θ)− 7

12

(r

a

)3

P3 +11

24

(r

a

)5

P5 −25

64

(r

a

)7

P7 + ...

(52)

16

1.2.3 Example: Potential of an Isolated Charge

Another method of finding the coefficients in the expansion makes use

of the fact that the expansion is unique. If, for example, we are able to

find the potential at fixed θ = θ0 for all r,

Φ(θ0, r) = g(r) =∞∑

l=0

(Alr

l +Blr−(l+1)

)Pl(cos θ0) (53)

then we can infer the form of the expansion by expanding g(r) in powers

of r and recognizing that the coefficient of rl must be Pl(cos θ0) times

a coefficient Al while the coefficient of (1/r)−(l+1) must be BlPl(cos θ0).

The most convenient value of θ0 to use is certainly 0 or π since we know

immediately the value of Pl(cos θ) in these instances.

Consider the following specific example: Suppose that there is a

charge q at a position x = az

(0,0,a)

in which case we know that the potential is

Φ(r, θ) =q√

r2 + a2 − 2ar cos θ. (54)

For θ = 0, we have simply φ(x) = q/|r − a|. At r < a in particular,

17

this function has a simple power series expansion,

Φ(r, 0) =q

a− r =q

a

1

1− r/a =q

a

1 +

r

a+r2

a2+r3

a3+ ...

. r < a

(55)

Hence, associating Pl with (r/a)l, we have

Φ(r, θ) =q

a

∞∑

l=0

(r

a

)lPl(cos θ); r < a (56)

the point is, the uniqueness of the expansion in terms of Legendre

polynomials tells us that this must be the solution. A similar expansion

done for r > a yields

Φ(r, θ) =q

r

∞∑

l=0

(a

r

)lPl(cos θ) r > a . (57)

There are two points that are worth making in connection with

these expansions. First, as stated earlier, there is a generating function

T (u, x) for the Legendre polynomials; see Eq. (29). We have just de-

rived it; that is, it is Φ, Eq. (54), equal to the sum in Eq. (56). Also,

we have obtained a convenient and useful expansion for the potential

of a point charge; in more general notation, we have derived

1

|x− x′| =∞∑

l=0

rl<rl+1>

Pl(cos γ) (58)

where γ is the angle between x and x′ while r< (r>) is the smaller

(larger) of |x| and |x′|.

18

1.3 Behavior of Fields in Conical Holes and Near Sharp Points

The field in the vicinity of the apex of a cone-shaped tip or depression

can also be investigated using the separation of variables method in

spherical coordinates. The solution for the potential is of the form, for

r small enough,

Φ(r, θ) ∼ rνPν(cos θ) (59)

where Pν(u) is a solution to the Legendre equation

d

du(1− u2)

dPνdu

+ ν(ν + 1)Pν = 0 (60)

with ν to be determined.z

P

r

θ

β

For the geometry shown, the solution must be well-behaved as θ → 0,

or u = cos θ → 1, but not necessarily as θ → π or u = −1. Introduce

19

the variable y ≡ 12(1− u) or u = 1− 2y; then Eq. (60) becomes

−1

2

d

dy

(1− (1− 2y)2

) (−1

2

dPνdy

)+ ν(ν + 1)Pν = 0 (61)

ord

dy

(y(1− y)

dPνdy

)+ ν(ν + 1)Pν = 0. (62)

Let us look once again for a solution in the form of a power series

expansion,

Pν = yα∞∑

j=0

ajyj, (63)

with 0 ≤ y ≤ y0 ≤ 1. Then

dPνdy

=∞∑

j=0

(α + j)ajyα+j−1, (64)

y(1− y)dPνdy

=∞∑

j=0

(α + j)aj(yα+j − yα+j+1), (65)

and

d

dy

(y(1− y)

dPνdy

)=∞∑

j=0

aj[(α + j)yα+j−1 − (α + j + 1)yα+j

](α + j).

(66)

Now combine these equations to find

∞∑

j=0

[aj(α + j)2yα+j−1 + aj ((α + j)(α + j + 1) + ν(ν + 1)) yα+j

]= 0

(67)

or, isolating individual powers of y,

a0α2yα−1 = 0 (68)

20

which implies that α = 0, and

aj+1 = ajj(j + 1)− ν(ν + 1)

(j + 1)2. (69)

If one lets ν = l, a non-negative integer, the result is just the Legendre

polynomials (no surprise), viewed as functions of y. More generally, for

any real ν > 0, one finds that the solutions are Legendre functions of

the first kind of order ν.

Pν =∞∑

j=0

aj(ν)yj, (70)

These are well-behaved (that is, they are not singular) functions of y

for y < 1 corresponding to u > −1 and are singular at y = 1.

For 1 > ν > 0, Pν(y) has a single zero; for 2 > ν > 1, Pν(y)

has two zeroes, etc. This is important because if we have a cone of

half-angle β with equipotential surfaces, we need ν to be such that

Pν ((1− cos β)/2) = 0. There will thus be a sequence of allowed values

of ν, which we designate by νk, k = 1, 2, 3,..., which are such that

yβ ≡ 12(1− cos β) ≡ kth zero of Pν.

The general solution at finite values of r, and including the point

r = 0 , is

Φ(r, θ) =∞∑

k=1

AkrνkPνk(cos θ). (71)

For small r, the leading term is the one with the smallest power of r,

that is the k = 1 term. Hence we may approximate the sum sufficiently

close to the origin by its leading term

Φ(r, θ) ≈ Arν1Pν1(cos θ). (72)

21

The dominant contribution to the electric field in this region comes

from this term; we have, by the usual E(x) = −∇Φ(x),

Er =dΦ

dr= −ν1Ar

ν1−1Pν1(cos θ) (73)

and

Eθ = −1

r

dΦ

dθ= A sin θ rν1−1 dPν1

(u)

du

∣∣∣∣∣∣cos θ

(74)

The behavior of ν1 as a function of β is shown below. For β less than

about 0.8π, one has1 ν1 ≈ 2.405β − 1

2 , while for β larger than about

the same number, ν1 ≈[2 ln

(2

π−β

)]−1. As β → π, ν1 → 0 and so

Er ∼ Eθ ∼ 1/r in this limit. The enhancement of a field near e.g., a

lightning rod is thus ∼ (R/δ) if R is the size of the system and δ is the

radius of curvature of the tip of the rod. Recall that in two dimensions

1This relation comes from study of the properties of the Legendre functions.

22

we found an enhancement of order (R/δ)1/2 The enhancement is much

more pronounced in three dimensions; a three dimensional tip is a much

sharper thing than an edge.

1.4 Associated Legendre Polynomials; Spherical Harmonics

Let us now return to the more general case of a solution to Laplace’s

equation (i.e. a potential) which depends on the azimuthal angle φ.

Then we must have the functions of φ eimφ, or, equivalently, sinmφ

and cosmφ, and the differential equation we have to face on the space

of θ isd

du(1− u2)

dP

du+

l(l + 1)− m2

1− u2

P = 0. (75)

The solutions are not finite polynomials in u in general but can be

expressed as infinite power series. They are only “well-behaved” on the

interval −1 ≤ u ≤ 1 when l ≥ |m|, with l an integer. Then there is just

one well-behaved solution which is known as the associated Legendre

function of degree l and order m. For m ≥ 0, the associated Legendre

function can be written in terms of the Legendre polynomial of the

same degree as

Pml (u) = (−1)m(1− u2)m/2

dm

dxm(Pl(u)) ; (76)

one can read all about this in Abramowitz and Stegun on pages 332 to

353. Making use of Rodrigues’ formula for the Legendre polynomials,

23

we see that

Pml (u) = (−1)m

(1− u2)m/2

2ll!

dl+m

dul+m[(u2 − 1)l]. (77)

This last formula is also valid for negative m 2; comparing the two

cases, one may see that

P−ml (u) = (−1)m(l −m)!

(l +m)!Pml (u). (78)

As for the Legendre polynomials, there is a generating function for

the associated Legendre functions as well as a variety of recurrence

relations. For example, a recurrence relation in degree is given by

(2l + 1)uPml (u) = (l −m+ 1)Pm

l+1(u) + (l +m)Pml−1(u) (79)

and one in order is

Pm+1l +

2mu√1− u2

Pml (u) + (l −m+ 1)(l +m)Pm−1

l (u) = 0 (80)

Out of all of this, what is of importance to us is that the product

(Arl +Br−l−1)Pml (cos θ)eimφ (81)

is a solution of the Laplace equation and that the set of functions

eimφPml (cos θ) with l = 0, 1, 2,..., and m = −l,−l + 1,...l − 1, l form

a complete orthogonal set on the two-dimensional domain 0 ≤ θ ≤ π

and 0 ≤ φ ≤ 2π. As usual, completeness is difficult to demonstrate

2That’s in part a matter of definition.

24

but orthogonality is quite straightforward using the formulae we have

already written down. Consider the integral

I =∫ 2π

0dφ

∫ π0dθ sin θe−imφPm

l (cos θ)eim′φPm′

l′ (cos θ) = 2πδmm′∫ 1

−1duPm

l (u)Pml′ (u)

(82)

Assume l′ ≥ l, m ≥ 0, and write Pml′ in terms of P−ml′ :

I = 2πδmm′(−1)m(l′ +m)!

(l′ −m)!

∫ 1

−1duP−ml′ (u)Pm

l (u)

= 2πδmm′(−1)m(l′ +m)!

(l −m)!

1

2l+l′l!l′!

∫ 1

−1du

dl′−m

dul′−m(u2 − 1)l

′ dl+m

dul+m(u2 − 1)l

= 2πδmm′(−1)l′ (l′ +m)!

(l′ −m)!

1

2l+l′l!l′!

∫ 1

−1du(u2 − 1)l

′ dl+l′

dul+l′(u2 − 1)l

= 2πδll′δmm′(l +m)!(2l)!(2l)!!

(l −m)!22l(l!)2(2l + 1)!!2 =

4π

2l + 1δll′δmm′

(l +m)!

(l −m)!.(83)

Thus we may construct an orthonormal set of functions on the surface

of the unit sphere; these are called spherical harmonics and are defined

as

Yl,m(θ, φ) ≡√√√√√

2l + 1

4π

(l −m)!

(l +m)!Pml (cos θ)eimφ, (84)

with m = −l,−l+1, ..., l−1,m and l = 0, 1, 2, .... These functions have

the property that

Y ∗l,m(θ, φ) = (−1)mYl,−m(θ, φ). (85)

The condition of orthonormality is

∫ 2π

0dφ

∫ π0

sin θdθY ∗l,m(θ, φ)Yl′m′,(θ, φ) = δll′δmm′. (86)

25

The completeness relation (not derived as usual) is

∞∑

l=0

m∑

m=−lY ∗l,m(θ′, φ′)Yl,m(θ, φ) = δ(φ− φ′)δ(cos θ − cos θ′). (87)

A general function g(θ, φ) is expanded in terms of the spherical har-

monics as

g(θ, φ) =∑

l,m

AlmYl,m(θ, φ) (88)

with

Alm =∫ 2π

0dφ

∫ π0dθ sin θg(θ, φ)Y ∗l,m(θ, φ). (89)

1.5 The Addition Theorem

In applications we will occasionally have need to know the function

Pl(cos γ) where γ is the angle between two vectors x and x′; it will

prove to be useful to be able to write this function in terms of the

variables θ, φ, θ′, and φ′.

x

y

z

θ

θ

φ

φ

γ

x

x’

’

’

It must be possible to do so in terms of any complete sets of functions of

these variables such as the spherical harmonics. In fact, the expansion

26

is

Pl(cos γ) =4π

2l + 1

l∑

m=−lY ∗l,m(θ′, φ′)Yl,m(θ, φ). (90)

We shall derive this expression as an example of the use of spherical

harmonics and their properties. First, let us set up a second coordinate

system rotated relative to the original one in such a way that its polar

axis lies along the direction of x′. In this system, the vector x has

components (rR, θR, φR).

x

y

zx

x’R

R

R

φR

γ = θR

Further, θR = γ. Next, we may regard Pl(cos γ) as a function of θ and

φ for fixed θ′ and φ′ and so can certainly expand it as

Pl(cos γ) =∞∑

l′=0

l∑

m′=−lAl′m′(θ

′, φ′)Yl′m′,(θ, φ). (91)

Similarly, in terms of spherical harmonics whose arguments are coordi-

nates in the rotated system, it is easy to see that

Pl(cos γ) =

√√√√ 4π

2l + 1Yl,0(θR, φR). (92)

27

Now, the spherical harmonics satisfy the differential equation

∇2Yl,m(θ, φ) +l(l + 1)

r2Yl,m(θ, φ) = 0 (93)

and they also satisfy this equation with variables θR, φR. But the

Laplacian operator ∇2 = ∇ · ∇ is a scalar object which is invariant

under coordinate rotations which means we can write it in the unrotated

frame while writing the spherical harmonic in the rotated frame:

∇2Yl,m(θR, φR) +l(l + 1)

r2Yl,m(θR, φR) = 0. (94)

Now recall that

Yl,0(θR, φR) =

√√√√2l + 1

4πPl(cos γ) =

√√√√2l + 1

4π

∞∑

l′=0

l′∑

m′=−l′Al′m′(θ

′, φ′)Yl′m′,(θ, φ).

(95)

If we plug this into the differential equation above, we obtain:

∞∑

l′=0

l′∑

m′=−l′Al′m′(θ

′, φ′)[∇2Yl′m′,(θ, φ) +l(l + 1)

r2Yl′m′,(θ, φ)] = 0 (96)

or

∞∑

l′=0

l′∑

m′=−l′Al′m′(θ

′, φ′)

−l

′(l′ + 1)

r2+l(l + 1)

r2

Yl′m′,(θ, φ) = 0. (97)

This equation can be true only if l = l′ or if Al′m′ = 0. Thus we have

demonstrated that Al′m′ = 0 for l′ 6= l, and the expansion of Pl reduces

to

Pl(cos γ) =l∑

m=−lAlm(θ′, φ′)Yl,m(θ, φ). (98)

28

The coefficients in this expansion are found in the usual way for an

orthogonal function expansion,

Alm =∫dΩPl(cos γ)Y ∗l,m(θ, φ) =

∫dΩRPl(cos θR)Y ∗l,m(θ, φ). (99)

Following the same line of reasoning, we may express√

4π/(2l + 1)Y ∗l,m(θ, φ)

as a sum of the form√√√√ 4π

2l + 1Y ∗l,m(θ, φ) =

l∑

m′=−lBlm′(m)Ylm′(θR, φR) (100)

where Bl0 in particular is

Bl0(m) =∫dΩR

√√√√ 4π

2l + 1Y ∗l,m(θ, φ)Y ∗l,0(θR, φR)

=∫dΩRY

∗l,m(θ, φ)Pl(cos θR) ≡ Alm. (101)

However, from Eq. (76), it is clear that when u = 1 Pml (u) = 0 when

m 6= 0,and Pml (u) = Pl(u) when m = 0, thus

√√√√ 4π

2l + 1Y ∗l,m(θ, φ)

∣∣∣θR=0

= Bl0(m)

√√√√2l + 1

4πPl(1) = Bl0(m)

√√√√2l + 1

4π,

(102)

so

Alm = Bl0(m) =4π

2l + 1Y ∗l,m(θ, φ)

∣∣∣θR=0

= Y ∗l,m(θ′, φ′)4π

2l + 1, (103)

where the last step follow since when θR = 0, x = x′. Thus,

Pl(cos γ) =4π

2l + 1

l∑

m=−lY ∗l,m(θ′, φ′)Yl,m(θ, φ). (104)

Thus ends our demonstration of the spherical harmonic addition theo-

rem.

29

An application if this theorem is that we can write 1|x−x′| as an ex-

pansion in spherical harmonics:

1

|x− x′| =∞∑

l=0

rl<rl+1>

Pl(cos γ) =∞∑

l=0

4π

2l + 1

rl<rl+1>

l∑

m=−l

[Y ∗l,m(θ′, φ′)Yl,m(θ, φ)

] .

(105)

This expansion is often useful when faced with common integrals in

electrostatics such as as∫d3x′ρ(x′)/|x− x′|.

1.6 Expansion of the Green’s Function in Spherical Harmon-

ics

More as an example of the use of the addition theorem than anything

else, let us devise an expansion for the Dirichlet Green’s function for

the region V bounded by r = a and r = b, a < b.

a

b

This function can be written as

G(x,x′) =1

|x− x′| + F (x,x′) (106)

where ∇2F = 0 in V. Thus it must be possible to write

G(x,x′) =∞∑

l=0

4π

2l + 1

rl<rl+1>

l∑

m=−lY ∗l,m(θ′, φ′)Yl,m(θ, φ)

30

+∞∑

l=0

l∑

m=−l

Alm

rl

bl+1+Blm

al

rl+1

Yl,m(θ, φ) (107)

where Alm and Blm can be functions of x′. The first term on the

right-hand side is 1|x−x′| and the second is a general solution of the

Laplace equation. The coefficients are determined by requiring that

the boundary conditions on G(x,x′) are satisfied (G(x,x′) = 0 for x

on one of the two bounding spherical surfaces). At r = a (r< = r = a

r> = r′), we have

0 =∑

l,m

4π

2l + 1

al

r′l+1Y ∗l,m(θ′, φ′) + Alm

al

bl+1+Blm

1

a

Yl,m(θ, φ) = 0,

(108)

from which it follows, using the orthogonality of the spherical harmon-

ics, that4π

2l + 1

al

r′l+1Y ∗l,m(θ′, φ′) + Alm

al

bl+1+Blm

1

a= 0. (109)

By similar means applied at r = b (r< = r′ r> = r = b), one may show

that4π

2l + 1

r′l

bl+1Y ∗l,m(θ′, φ′) + Alm

1

b+Blm

al

bl+1= 0. (110)

These present us with two linear equations that may be solved for Alm

and Blm; the solutions are

Alm =4π

(2l + 1)Y ∗l,m(θ′, φ′)

a

2l+1

blr′l+1− r′l

bl

/1−

(a

b

)2l+1 (111)

and

Blm =4π

2l + 1Y ∗l,m(θ′, φ′)

a

l+1r′l

b2l+1− al+1

r′l+1

/1−

(a

b

)2l+1 . (112)

31

From these and the expansion Eq. (107), we find that the Green’s

function is

G(x,x′) =∑

l,m

4π

(2l + 1)[1−

(ab

)2l+1]Y ∗l,m(θ′, φ′)Yl,m(θ, φ)

1−

(a

b

)2l+1 rl<rl+1>

+

(a

b

)2l+1 rl

r′l+1− r′lrl

b2l+1+

(a

b

)2l+1 r′l

rl+1− a2l+1

rl+1r′l+1

.(113)

This result, if we can call it that, can be written in a somewhat more

compact form by factoring the quantity .... Suppose that r> = r′

and r< = r; then

... =

rl − a2l+1

rl+1

1

r′l+1− r′l

b2l+1

≡rl< −

a2l+1

rl+1<

1

rl+1>

− rl>b2l+1

. (114)

If, on the other hand, r> = r and r< = r′, then

... =

r′l − a2l+1

r′l+1

1

rl+1− rl

b2l+1

≡

rl< −

a2l+1

rl+1<

1

rl+1>

− rl>b2l+1

.

(115)

Comparing these results, we see that we may in general write

G(x,x′) =∑

l,m

4π/(2l + 1)

1−(ab

)2l+1 Y∗l,m(θ′, φ′)Yl,m(θ, φ)

rl< −

a2l+1

rl+1<

1

rl+1>

− rl>b2l+1

.

(116)

Notice that the Green’s functions for the interior of a sphere of radius

b and for the exterior of a sphere of radius a are easily obtained by

taking the limits a → 0 and b → ∞, respectively. In the former case,

32

for example, one finds

G(x,x′) =∑

l,m

4π

2l + 1Y ∗l,m(θ′, φ′)Yl,m(θ, φ)

rl<rl+1>

− rl<rl>

b2l+1

=1

|x− x′| −∑

l,m

4π

2l + 1Y ∗l,m(θ′, φ′)Yl,m(θ, φ)

b

r′rl

(b2/r′)l+1

=1

|x− x′| −b/r′

|x− x′R|(117)

where x′R = (b2/r′, θ′, φ′) in spherical coordinates.

2 Laplace Equation in Cylindrical Coordinates; Bessel

Functions

In cylindrical coordinates the Laplacian is

∇2 =∂2

∂ρ2+

1

ρ

∂

∂ρ+

1

ρ2

∂2

∂φ2+

∂2

∂z2. (118)

We once again look for solutions of the Laplace equation in the form

of products of functions of a single variable,

Φ(x) = R(ρ)Q(φ)Z(z); (119)

Following the usual procedure (substitute into the Laplace equation;

divide by appropriate functions to obtain terms which appear to depend

on a single variable; argue that such terms must be constants; etc.), we

wind up with the following three ordinary differential equations:

d2Q

dφ2+ ν2Q = 0, ν = 0,±1,±2,..., (120)

33

d2Z

dz2− k2Z = 0, (121)

andd2R

dρ2+

1

ρ

dR

dρ+

k2 − ν2

ρ2

R = 0 (122)

where the value of k is yet to be determined, and ν is determined as

indicated by the same argument as in the case of spherical coordinates;

the functions Q(φ) are the same as in spherical coordinates also, Q(φ) ∼eiνφ.

The choice of k is specified by the sort of boundary conditions one

has. One could imagine having to satisfy quite arbitrary conditions on

an end face z = c where c is constant; alternatively, one may have to

fit some function on a side wall ρ = c. In the former case, one wants

to have functions of ρ which form a complete set on an appropriate

interval of ρ; and in the latter case, one wants functions of z to form a

complete set on some interval of z; in both cases we will need a complete

set of functions of φ, which we have. Now, looking at the equations for

R and Z, we can see that the latter function in particular is going to be

simple exponentials of kz; for k real, these do not form a complete set;

for k imaginary, they are sines and cosines and can form a complete

set. We may not recognize it yet, but a similar thing happens to R; for

k imaginary, it is roughly exponential in character and we cannot get a

complete set of functions in this way. But for k real, the functions R are

oscillatory (although not sines and cosines) and can form a complete

34

set.

k Z(z) R(ρ)

real incomplete (e±kz) complete (oscillatory)

imaginary complete (ei±|k|z) incomplete

The functions of z in either case (k real or imaginary) are familiar to

us and do not require further discussion. The functions of R, although

probably known to all of us at least vaguely, are much less familiar

so we will spend some time presenting their most important, to us,

properties. Let’s start by defining a dimension-free variable x = kρ.

Then Eq. (122) becomes

k2d2R

dx2+ k2 1

x

dR

dx+ k2

1− ν2

x2

R = 0 (123)

ord2R

dx2+

1

x

dR

dx+

1− ν2

x2

R = 0 (124)

which is Bessel’s Equation. Its solutions are Bessel functions of order

ν. In our particular case, ν is an integer, although this need not be

true in general. If k is imaginary, then x is imaginary, so we must deal

with Bessel functions of imaginary argument; viewed as functions of a

real variable |x|, these are known as modified Bessel functions.

For a given ν, there are two linearly independent solutions of Bessel’s

equation. Their choice is somewhat arbitrary since any linear combi-

nation of them is also a solution. One possible and common way to

35

choose them is as follows:

Jν(x) =

(x

2

)ν ∞∑

j=0

(−1)j(x/2)2j

j!Γ(j + ν + 1)(125)

and

J−ν(x) =

(x

2

)−ν ∞∑

j=0

(−1)j(x/2)2j

j!Γ(j − ν + 1)(126)

where

Γ(z) ≡∫ ∞

0dt tz−1e−t (127)

is the gamma function which, for z a real, positive integer n, is Γ(n) =

(n− 1)!. For z = 0 or a negative integer, it is singular.

The two Bessel functions introduced above are linearly independent

solutions of Bessel’s equation so long as ν is not an integer. It is easy to

verify that they are solutions by direct substitution into the differential

equation. If, however, ν is an integer, they become identical (aside from

a possible sign difference) and so do not provide us with everything we

need in this case, which is the important one for us. Another function,

which is a solution and which is linearly independent of either of the

two solutions introduced above (taken one at a time) is given by

Nν(x) ≡ Jν(x) cos(νπ)− J−ν(x)

sin(νπ). (128)

This is the Neumann function; it is also called the Bessel function of

the second kind of order ν. 3

3Why does this work? Consider the limit ν → m by La Hopital’s rule. To formally show that

Nm and Jm are independent, one must calculate the Wronskian and show that W [Nm, Jm] 6= 0.Note

that in Abramowitz and Stegun’s book, this function is written as Yν(x); see p. 358.

36

For ν = n, a non-negative integer, the Neumann function has a series

representation which is

Nn(x) ≡ −1

π

(x

2

)−n n−1∑

j=0

(n− j − 1)!

j!

(x

2

)2j

+2

πln(x/2)Jn(x)

−1

π

(x

2

)n ∞∑

j=0

[ψ(j + 1) + ψ(n+ j + 1)]

(x/2)2j(−1)j

j!(n+ j)!

(129)

where ψ(y) = d(ln Γ(y))/dy is known as the digamma or psi function.

Finally, for some purposes it is more useful to use Bessel functions

of the third kind, also called Hankel functions; these are given by

H(1)ν = Jν(x) + iNν(x) (130)

and

H(2)ν = Jν(x)− iNν(x). (131)

It is not easy to see what are the properties of the various kinds of

Bessel functions from the expansions we have written down so far. As it

turns out, their behavior is really quite simple; many of the important

features are laid bare by their behavior at small and large arguments.

For x << 1 and real, non-negative ν, one finds

Jν(x) =1

Γ(ν + 1)

(x

2

)ν[1−O(x2)] (132)

and

Nν(x) =

2π [ln(x/2) + 0.5772 + ...] ν = 0

−Γ(ν)π

(2x

)νν 6= 0

(133)

37

For x >>> 1, ν,

Jν(x) ∼√√√√ 2

πxcos

(x− νπ

2− π

4

)(134)

and

Nν(x) ∼√√√√ 2

πxsin

(x− νπ

2− π

4

). (135)

0 5 10 15x=kρ

-3

-2

-1

0

1

Ord

inar

y B

esse

l’s

func

tions

J0(x)J1(x)N0(x)N1(x)

Notice in particular that as x → 0, the Bessel functions of the first

kind are well-behaved (finite) whereas the Neumann functions are sin-

gular; N0(x) has only a logarithmic singularity while the higher-order

functions are progressively more singular. At large (real) argument, on

the other hand, both Jν(x) and Nν(x) are finite and oscillatory. Hence,

the Bessel functions (by which we mean the Jν(x)) of non-negative or-

der are allowable as solutions of the Laplace equation at all values of

ρ; the Neumann functions, on the other hand, are not allowable on a

domain which includes the point ρ = 0.

38

Bessel functions of all kinds satisfy certain recurrence relations. It

is a straightforward if tedious matter to show by direct substitution of

the series expansions that they obey the following:

Ων+1 −2ν

xΩν + Ων−1 = 0 (136)

and

Ων+1 + 2dΩν

dx− Ων−1 = 0. (137)

By taking the sum and difference of these relations, we find also

Ων±1 =ν

xΩν ∓

dΩν

dx. (138)

These are valid for all three kinds of Bessel functions.

The Bessel function Jν(x) can form a complete orthogonal set on

an interval 0 ≤ x ≤ x0 in much the same way as the sine function

sin(nπx/x0) does (Note that x0 is a zero of the sine function.) Similarly,

let us denote the nth zero of Jν(x) by xνn and then form the functions

Jν(xνny), with n = 1,2,... . Then it turns out that for fixed ν, these

functions provide a complete orthogonal set on the interval 0 ≤ y ≤ 1.

As usual, we shall not demonstrate completeness. Orthogonality can

be demonstrated by making use of the Bessel equation and recurrence

relations. It is a useful exercise to do so. Start from the Bessel equation

for Jν(xy),

1

y

d

dy

ydJν(xy)

dy

+

x2 − ν2

y2

Jν(xy) = 0 (139)

39

Multiply this equation by yJν(x′y) and integrate from 0 to 1:

∫ 1

0dy

Jν(x

′y)d

dy

ydJν(xy)

dy

+ yJν(x

′y)

x2 − ν2

y2

Jν(xy)

= 0

(140)

or if we integrate the first term by parts:

Jν(x′y)y

dJν(xy)

dy

∣∣∣∣∣∣

1

0

−∫ 1

0dydJν(x

′y)

dyydJν(xy)

dy+

∫ 1

0dyyJν(x

′y)Jν(xy)

x2 − ν2

y2

= 0 (141)

Similarly, if we start from the differential equation for Jν(x′y) and per-

form the same manipulations, we find the same equation with x and x′

interchanged. Subtract the second equation from the first to find

Jν(x′)x

dJν(u)

du

∣∣∣∣∣∣x

−Jν(x)x′dJν(u)

du

∣∣∣∣∣∣x′

+(x2−x′2)∫ 1

0ydyJν(xy)Jν(x

′y) = 0

(142)

If we let x = xνn and x′ = xνn′, two distinct zeros of the Bessel function,

then the integrated terms vanish and we may conclude that the Bessel

functions Jν(xνny) and Jν(xνn′y) are orthogonal when integrated over

y from 0 to one, provided a factor of y is included in the integrand.

We still have to determine normalization in the case n = n′. In the

preceding equation, let x′ = xνn and rearrange the terms to have

−Jν(x)xνndJν(u)

du

∣∣∣∣∣∣xνn

= −∫ 1

0dy y(x2 − x2

νn)Jν(xνny)Jν(xy), (143)

or∫ 1

0dy yJν(xνny)Jν(xy) =

Jν(x)xνn(dJν(u)/du)|xνnx2 − x2

νn

. (144)

40

Use L’Hopital’s Rule to evaluate the limit of this expression as x→ xνn:

∫ 1

0dy y[Jν(xνny)]2 =

1

2

dJν(x)

dx

∣∣∣∣∣∣xνn

dJν(u)

du

∣∣∣∣∣∣xνn

=1

2[J ′ν(xνn)]2, (145)

where the prime ’ denotes a derivative with respect to argument. Now

employ the recurrence relation

xJ ′ν(x) = νJν(x)− xJν+1(x) (146)

to find J ′ν(xνn) = −Jν+1(xνn), from which the normalization integral

becomes∫ 1

0dy y[Jν(xνny)]2 =

1

2[Jν+1(xνn)]2 (147)

The expansion of an arbitrary function of ρ on the interval 0 ≤ ρ ≤ a

may be written as

f(ρ) =∞∑

n=1

AnJν(ρxνn/a) (148)

with coefficients which may be determined from the orthonormalization

properties of the basis functions as

An =

∫ a0 ρdρf(ρ)Jν(ρxνn/a)

a2

2 [Jν+1(xνn)]2. (149)

This type of expansion is termed a Fourier-Bessel series. The com-

pleteness relation for the basis functions is

∞∑

n=1

Jν(ρxνn/a)Jν(ρ′xνn/a)

(a2/2)[Jν+1(xνn)]2= δ(ρ2/2− ρ′2/2) ≡ 1

ρδ(ρ− ρ′). (150)

It is also of importance to consider the case of imaginary k, k =

iκ with real κ. Then the functions of z are oscillatory, being of the

41

form Z(z) ∼ e±iκz, and the functions of ρ will be Bessel functions of

imaginary argument, e.g., Jν(iκρ). For given ν2, there are two linearly

independent solutions which are conventionally chosen to be Jν and

H(1)ν , the reason being that they have particularly simple behaviors

at large and small arguments. Let us introduce the modified Bessel

functions Iν(x) and Kν(x),

Iν(x) ≡ i−νJν(ix) (151)

Kν(x) ≡ π

2iν+1H(1)

ν (ix). (152)

These have the forms at small argument, x << 1,

Iν(x) =1

Γ(ν + 1)

(x

2

)ν(153)

and

Kν(x) =

− ln(x/2) + 0.5772 + ... ν = 0

Γ(ν)2

(2x

)νν 6= 0

(154)

while at large argument, x >>> 1, ν

Iν(x) =1√2πx

ex[1 +O

(1

x

)](155)

and

Kν(x) =

√π

2xe−x

[1 +O

(1

x

)]. (156)

42

From these equations we see that Iν(x) is well-behaved for x < ∞,

corresponding to ρ < ∞, while Kν(x) is well-behaved for x > 0 or

ρ > 0, from which we can decide which function(s) to use in expanding

any given potential problem.

Let’s look at some examples of expansions in cylindrical coordinates.

2.1 Example I

Consider a charge-free right-circular cylinder bounded by S given by

ρ = a, z = 0, and z = c. Let Φ(x) be zero on S except for the top face

z = c where Φ(ρ, φ, c) = V (ρ, φ) with V given.

43

x

y

z

ac

Φ = 0

Φ = V( )ρ,φ

For this distribution of boundary potential, we need a complete set

of functions of the space 0 ≤ φ ≤ 2π and 0 ≤ ρ ≤ a. Thus we take (k

real) Z(z) to be damped exponentials (sinh and cosh) and R(ρ) to be

ordinary Bessels functions.

Φ ∼ eimφ (AJm(kρ) +BNm(kρ)) (cos(kz)± sin(kz)) (157)

It will be convenient if each of these functions is equal to zero when

ρ = a and also if each one is zero when z = 0. With just a little

thought,

1. No Neumann functions Nm since they diverge at ρ = 0

2. No cosh since it is finite at z = 0, and hence would not satisfy the

B.C.

3. Since Jm and J−m are not independent, use J|m|.

we realize that we want to use the hyperbolic sine function of z and

the Bessel function of the first kind for R. Our expansion is thus of the

44

form

Φ(ρ, φ, z) =∞∑

m=−∞

∞∑

n=1

AmneimφJ|m|(xmnρ/a) sinh(xmnz/a) (158)

where xmn is the nth zero of J|m|(x). Each term in the sum is itself

a solution of the Laplace equation; each one satisfies the boundary

conditions on z = 0 and ρ = a, and, for given n, we have a complete

set of functions of φ while for given m, we have a complete set of

functions of ρ.

The coefficients in the expansion are determined from the condition

that Φ reduce to the given potential V on the top face of the cylinder.

Making use of the orthogonality of the basis functions of both φ and ρ,

we have∫ a

0ρdρ

∫ 2π

0dφV (ρ, φ)J|m|(xmnρ/a)e−imφ (159)

= Amn2π(a2/2)[J|m|+1(xmn)]2 sinh(xmnc/a)

or

Amn =

∫ 2π0 dφ

∫ a0 ρdρe

−imφJ|m|(xmnρ/a)V (ρ, φ)

πa2 sinh(xmnc/a)[J|m|+1(xmn)]2. (160)

For any given function V , one may now attempt to complete the inte-

grals.

2.2 Example II

Consider the same geometry as in the first example but now with

boundary condition Φ = 0 on the constant-z faces and some given

value V (φ, z) on the surface at ρ = a.

45

x

y

z

ac

Φ = 0

Φ = 0

= V( ,z)Φ φ

For this system we need a complete set of functions on the domain

0 ≤ z ≤ c and 0 ≤ φ ≤ 2π which means picking k imaginary, k = iκ.

The appropriate functions of z are sin and cos, and the appropriate

functions of ρ are the modified Bessels Functions.

Φ ∼ eimφ (sin(kz)± cos(kz)) (AI(kρ) +BK(kρ)) (161)

1. We may eliminate the K modified Bessels functions since they

diverge when ρ→ 0.

2. Since Im is not independent of I−m, we use I|m|.

3. The cos function of z cannot be zero at both z = 0 and z = c, and

so may be eliminated.

4. Take k = nπ/c so that sin(nπz/c)|z=c = 0

46

Thus the expansion is

Φ(ρ, φ, z) =∞∑

m=−∞

∞∑

n=1

Amneimφ sin(nπz/c)I|m|(nπρ/c) (162)

with coefficients given by

Amn =

∫ c0 dz

∫ 2π0 dφe−imφ sin(nπz/c)V (φ, z)

πcI|m|(nπa/c)(163)

2.3 B.V.P. on Large Cylinders

By applying the same considerations, one may solve other boundary-

value problems on cylinders. A case of special interest, and requiring

special treatment, is one in which a→∞; then kνn ≡ xνn/a becomes a

continuous variable and instead of a Fourier-Bessel series, we come up

with an integral. The orthogonality condition is

∫ ∞0xdxJm(kx)Jm(k′x) =

1

kδ(k − k′) (164)

and the completeness relation is the same, with different names for the

variables,∫ ∞

0kdkJm(kx)Jm(kx′) =

1

xδ(x− x′). (165)

To see how this comes to be, consider the completeness relation on a

finite interval,

∞∑

n=1

Jm(xmnρ/a)Jm(xmnρ′/a)

(a2/2)[Jm+1(xmn)]2=

1

ρδ(ρ− ρ′) (166)

and then let a→∞, defining xmn/a as k while noting that the interval

between roots of the Bessel function at large argument is π. Also, the

47

asymptotic form of the Bessel function, valid at large argument, is

Jm+1(xmn) ∼√√√√ 2

πxmncos[nπ + (m− 1/2)π/2− (m+ 1)π/2− π/4]

=

√√√√ 2

πxmncos[(n− 1)π] = −(−1)n

√√√√ 2

πxmn.(167)

Using this substitution in the closure relation for the finite interval

and taking the limit of large a, one finds that the sum becomes the

integral, Eq. (165). Of course, this is not a rigorous derivation because

the asymptotic expression is not arbitrarily accurate for all roots.

2.4 Green’s Function Expansion in Cylindrical Coordinates

We can expand the Dirichlet Green’s function in cylindrical coordinates

in much the same manner as we did in spherical coordinates. We shall

go through the derivation to expose a somewhat different approach from

what we employed in the latter case. We consider a domain between

two infinitely long right-circular cylindrical surfaces. Then G must

vanish on these surfaces and it must also satisfy a Poisson equation,

∇2G(x,x′) = −4πδ(ρ2/2− ρ′2/2)δ(φ− φ′)δ(z − z′)

= −4π

ρδ(ρ− ρ′)δ(φ− φ′)δ(z − z′). (168)

Let us write the delta functions of φ and z using closure relations,

δ(z − z′) =1

2π

∫ ∞−∞

dkeik(z−z′) =1

π

∫ ∞0dk cos[k(z − z′)] (169)

48

and

δ(φ− φ′) =1

2π

∞∑

m=−∞eim(φ−φ′). (170)

Similarly, expand the φ-dependence and z-dependence of G using the

same basis functions,

G(x,x′) =∞∑

m=−∞

∫ ∞0

dk

2π2gm(k, ρ, ρ′)eim(φ−φ′) cos[k(z − z′)]. (171)

Now operate on this expansion with the Laplacian using cylindrical

coordinates:

∇2G(x,x′) =∞∑

m=−∞

∫ ∞0

dk

2π2

d2

dρ2+

1

ρ

d

dρ−m

2

ρ2+ k2

gme

im(φ−φ′) cos[k(z − z′)]

= −4π

ρδ(ρ− ρ′)

∞∑

m=−∞

∫ ∞0

dk

2π2eim(φ−φ′) cos[k(z − z′)].(172)

Multiply by members of the basis sets, i.e., e−im′(φ−φ′) and cos[k′(z−z′)]

and integrate over the appropriate intervals of φ− φ′ and z− z′ to find

a differential equation for gm,

d2gmdρ2

+1

ρ

dgmdρ−k2 +

m2

ρ2

gm = −4π

ρδ(ρ− ρ′). (173)

For ρ 6= ρ′, this is Bessel’s equation with solutions (viewed as functions

or ρ) which are Bessel functions of imaginary argument, or, as we have

described, modified Bessel functions of argument kρ. Because of the

delta function inhomogeneous term, the solution for ρ < ρ′ is different

from the solution for ρ > ρ′. Hence we may write that, for ρ < ρ′,

gm(k, ρ, ρ′) = A<(ρ′)Km(kρ) +B<(ρ′)Im(kρ), (174)

49

and, for ρ > ρ′,

gm(k, ρ, ρ′) = A>(ρ′)Km(kρ) +B>(ρ′)Im(kρ). (175)

The various coefficients are functions of ρ′ and must in fact be linear

combinations of Km(ρ′) and Im(ρ′) because G(x,x′) is also a solution

of ∇′2G(x,x′) = 0 when ρ 6= ρ′; another way to see this same point is

to recall that G(x,x′) = G(x′,x). Finally, the coefficients are further

constrained by the condition that the Green’s function must vanish

when ρ becomes equal to the radius of either the inner or outer cylinder.

Let us at this point restrict our attention to a special (and simple)

limiting case which is the infinite space. The radius of the inner cylinder

is 0 and that of the outer one becomes infinite in this limit. Then we

have to have a function gm which remains finite as ρ → 0 which can

only be Im(kρ); also, we must have gm vanish as ρ → ∞, which can

only be Km(kρ). Thus we have

gm(k, ρ, ρ′) =

A<(ρ′)Im(kρ) ρ < ρ′

A>(ρ′)Km(kρ) ρ > ρ′.(176)

The symmetry condition on G tells us that A<(ρ′) = AKm(kρ′) while

A>(ρ′) = AIm(kρ′). All of these conditions are included in the state-

ment

gm(k, ρ, ρ′) = AIm(kρ<)Km(kρ>) (177)

where ρ< (ρ>) is the smaller (larger) of ρ and ρ′.

50

The remaining constant in the determination of gm can be found

from the required normalization of G. Let us integrate Eq. (173) across

the point ρ = ρ′:

∫ ρ′+ερ′−ε

dρ

d2

dρ2+

1

ρ

d

dρ−k2 +

m2

ρ2

gm = −4π

ρ′(178)

If we take the limit of ε→ 0 and realize that gm is continuous while its

first derivative is not, we find that this equation gives

limε→0

dgmdρ

∣∣∣∣∣

ρ′+ε

ρ′−ε= A

Im(kρ′)k

dKm(x)

dx

∣∣∣∣∣∣kρ′−Km(kρ′)k

dIm(x)

dx

∣∣∣∣∣∣kρ′

= −4π

ρ′,

(179)

or

A[Im(x)K ′m(x)−Km(x)I ′m(x)] = −4π

x; (180)

the primes denote derivatives with respect to the argument x. The

quantity in [...] here is the Wronskian of Im and Km. One may learn

by consulting, e.g., the section on Bessel functions in Abramowitz and

Stegun, that Bessel functions have simple Wronskians:

Im(x)K ′m(x)−Km(x)I ′m(x) ≡ W [Im(x), Km(x)] = −1

x. (181)

Comparison of the two preceding equations leads one to conclude that

A = 4π. Hence our expansion of G(x,x′), which is just 1/|x− x′|, is

G(x,x′) =1

|x− x′| =2

π

∞∑

m=−∞

∫ ∞0dkeim(φ−φ′) cos[k(z−z′)]Im(kρ<)Km(kρ>)

(182)

51

which may also be written entirely in terms of real functions as

1

|x− x′| =4

π

∫ ∞0dk cos[k(z − z′)]

×

1

2I0(kρ<)K0(kρ>) +

∞∑

m=1

cos[m(φ− φ′)]Im(kρ<)Km(kρ>)

. (183)

This turns out to be a useful expansion of 1/|x− x′|; it also provides a

starting point for the derivation of some other equally useful expansions.

For example, if we let x′ = 0, then ρ< = 0 and all Im vanish except for

m = 0, while 1/|x − x′| = 1/|x| = 1/√ρ2 + z2, so we find, using also

I0(0) = 1,1√

ρ2 + z2=

2

π

∫ ∞0dk cos(kz)K0(kρ). (184)

Other useful identities may be obtained.

52